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3 years ago

A car is traveling at a speed of 20m/s and has a mass of 1200kg how much kinetic energy does it have

Physics

1 answer:

Oxana [17]3 years ago

4 0

Answer:

Explanation:

The formula for Kinetic Energy is

$KE=\frac{1}{2}mv^2$ . Filling in:

$KE=\frac{1}{2}(1200)(20)^2$ It looks like we only need 1 significant digit here but I'll give you 2 and you can round how you want.

KE = 2.4 × 10⁵ J

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A meter stick is free to rotate about an axis through one of its end. Find the force F needed to balance this meter stick if the

Mumz [18]

Answer:

Explanation:

Component of force perpendicular to stick

= F Sin 60°

=√3 / 2 F.

Taking torque about the other end

= √3 / 2 F x 1 Nm

Weight of stick = 60 gm

= 60 x 10⁻³ kg

= 60 x 10⁻³ x 9.8 N

= .588 N

This weight will act from the middle point of stick so torque about the

other end

= .588 x 1 Nm

Balancing these two torques we have

.588 = √3 /2 F

$F=\frac{2\times0.588}{\sqrt{3} }$

F = 0.679 N

6 0

3 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0

3 years ago

______ is the total distance traveled divided by the total time of travel.

sweet-ann [11.9K]

Average speed is the answer

4 0

3 years ago

You wish to move a heavy box on a rough floor. The coefficient of kinetic friction between the box and the floor is 0.9. What is

Maksim231197 [3]

Answer:

Explanation:

5 0

3 years ago

A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c

icang [17]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

$W = F x = \mu N x = \mu\ m\ g\ x$

$W = 150 \times 6$

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

$\mu = \frac {F}{m\timesg}$

$= \frac{150}{40\times 9.8}$

= .383

We simply applied the above formulas so that each one part could calculate

3 0

4 years ago